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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $-\large\frac{x}{2}$$<\sin^{-1}x<\large\frac{x}{2}$ then $\tan(\sin^{-1}x)$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{x}{1-x^2}&(b)\;\large\frac{x}{1+x^2}\\(c)\;\large\frac{x}{\sqrt{1-x^2}}&(d)\;\large\frac{1}{\sqrt{1-x^2}}\end{array}$

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$\tan(\sin^{-1}x)=\tan\big(\tan^{-1}\large\frac{x}{\sqrt{1-x^2}}\big)$$\;x\in (-1,1)$
$\Rightarrow \large\frac{x}{\sqrt{1-x^2}}$
Hence (d) is the correct option.
answered Nov 15, 2013 by sreemathi.v
 

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